Consider the contour γ given by \(\rm \gamma (\theta)=\left\{\begin{matrix}e^{2i \theta}&for\ \theta \in [0, \pi /2]\\\ 1+2e^{2i\theta}&for\ \theta \in [\pi/2, 3\pi/2]\\\ e^{2i\theta}&for\ \theta \in [3\pi/2, 2\pi]\end{matrix}\right.\)

Then what is the value of \(\rm \int_{\gamma}\frac{dz}{z(z-2)}\)?

Concept:

Cauchy Residue Theorem:

Let \( f(z) \) be a function that is analytic inside and on a simple closed contour C, except for a finite number of

isolated singularities (poles) inside C. If \( f(z)\) has isolated singularities at  \(z_1, z_2, \dots, z_n \) inside C, then the integral of \( f(z) \) around  C is given by

\(\oint_C f(z) \, dz = 2\pi i \sum_{k=1}^{n} \text{Res}(f, z_k) \)

Explanation:

\(\int_{\gamma} \frac{dx}{z(z - 2)}\), where the contour \(\gamma(\theta) \) is piecewise defined as:

\(\gamma(\theta) = \begin{cases} e^{2i\theta} & \text{for } \theta \in [0, \frac{\pi}{2}] \\ 1 + 2e^{2i\theta} & \text{for } \theta \in [\frac{\pi}{2}, \frac{3\pi}{2}] \\ e^{2i\theta} & \text{for } \theta \in [\frac{3\pi}{2}, 2\pi] \end{cases}\)

We will apply the Cauchy Residue Theorem to solve this integral. The integrand is

\(f(z) = \frac{1}{z(z - 2)}\)

This function has two singularities at \( z = 0 \)  and \(z = 2 \) . If the contour \(\gamma\) encloses these singularities, the value of the integral is determined by the residues of the function at these points.

Here winding number of z = 0 is 2 and winding number of z = 2 is 1.

Residue at \( z = 0 \) :

The residue at \( z = 0 \) is found by multiplying \(f(z) \) by \(z\) and taking the limit as  \(z \to 0 \):

\(\text{Res}(f, 0) = \lim_{z \to 0} z \cdot \frac{1}{z(z - 2)} = \frac{1}{-2} = -\frac{1}{2}\)

Residue at \( z = 2 \):

The residue at \( z = 2 \) is found similarly,

\(\text{Res}(f, 2) = \lim_{z \to 2} (z - 2) \cdot \frac{1}{z(z - 2)} = \frac{1}{2}\)

Applying the Residue Theorem:

Since the contour \( \gamma \) encloses both singularities at \( z = 0 \) and \( z = 2 \) , the value of the integral is

I = \(2\pi i \times (\text{sum of residues}) \)

  = \( 2\pi i \times \left(-\frac{1}{2}\times 2 + \frac{1}{2}\times1\right)\)

  = \(2\pi i(-\frac12)\) = -πi

Hence, the correct option is (3).